离散化都忘记怎么写了 注意两个端点 离散化后用线段树更新区间 混色为-1 黑为2 白为1 因为N不大 最后直接循环标记这一段的颜色查找
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define N 100010 8 #define LL long long 9 struct node 10 { 11 int d,id; 12 char c; 13 }li[N<<1]; 14 int q[N][3],pp[N]; 15 int s[N<<2],cnt[N],cc[N]; 16 void build(int l,int r,int w) 17 { 18 s[w] = 1; 19 if(l==r-1) 20 { 21 return ; 22 } 23 int m = (l+r)>>1; 24 build(l,m,w<<1); 25 build(m,r,w<<1|1); 26 } 27 void update(int a,int b,int d,int l,int r,int w) 28 { 29 if(a<=l&&b>=r) 30 { 31 s[w] = d; 32 return ; 33 } 34 if(l==r-1) 35 return ; 36 if(s[w]>0&&s[w]!=d) 37 { 38 s[w<<1] = s[w<<1|1] = s[w]; 39 s[w] = -1; 40 } 41 int m = (l+r)>>1; 42 if(a<=m) 43 update(a,b,d,l,m,w<<1); 44 if(b>m) 45 update(a,b,d,m,r,w<<1|1); 46 } 47 void query(int l,int r,int w) 48 { 49 int i; 50 if(s[w]>0) 51 { 52 for(i = l ; i < r; i++) 53 cnt[i] = s[w]; 54 return ; 55 } 56 if(l==r-1) 57 return ; 58 int m = (l+r)>>1; 59 query(l,m,w<<1); 60 query(m,r,w<<1|1); 61 } 62 bool cmp(node a,node b) 63 { 64 return a.d<b.d; 65 } 66 int main() 67 { 68 int i,n; 69 char c[10]; 70 scanf("%d",&n); 71 for(i = 0 ; i < n ; i++) 72 { 73 scanf("%d%d%s",&q[i][0],&q[i][1],c); 74 if(c[0]=='b') 75 cc[i+1] = 2; 76 else 77 cc[i+1] = 1; 78 li[2*i].d = q[i][0]; 79 li[2*i].id = -(i+1); 80 li[2*i+1].d = q[i][1]; 81 li[2*i+1].id = i+1; 82 } 83 sort(li,li+2*n,cmp); 84 int tt = li[0].d,g=1; 85 for(i = 0 ; i < 2*n ; i++) 86 { 87 if(li[i].d!=tt) 88 { 89 g++; 90 tt = li[i].d; 91 } 92 if(li[i].id<0) 93 { 94 q[-li[i].id][0] = g; 95 pp[g] = li[i].d; 96 } 97 else 98 { 99 q[li[i].id][1] = g; 100 pp[g] = li[i].d; 101 } 102 } 103 build(0,g,1); 104 for(i = 0; i <= g ; i++) 105 cnt[i] = 1; 106 for(i = 1 ; i <= n ; i++) 107 { 108 update(q[i][0],q[i][1],cc[i],0,g,1); 109 } 110 query(0,g,1); 111 int maxz=0,ss,ee,ts,te; 112 cnt[0] = 1; 113 cnt[g+1] = 1; 114 pp[g+1] = 1000000000; 115 for(i = 0 ; i <= g ; i++) 116 { 117 if(cnt[i]!=1) 118 continue; 119 ss = pp[i]; 120 while(cnt[i]==1&&i<=g) 121 i++; 122 ee = pp[i]; 123 if(ee-ss+1>maxz) 124 { 125 ts = ss; 126 te = ee; 127 maxz = ee-ss+1; 128 } 129 } 130 printf("%d %d ",ts,te); 131 return 0; 132 }